Resolver sina=sqrt{b}(1-sin^2a)

$$\sin a = \sqrt { b } ( 1 - \sin ^ { 2 } a )$$

Ebatzi: b

$$b=\frac{\left(\tan(a)\right)^{2}}{\left(\cos(a)\right)^{2}}$$

$$\exists n_{1}\in \mathrm{Z}\text{ : }\left(\left(a\geq 2\pi n_{1}\text{ and }a<2\pi n_{1}+\frac{\pi }{2}\right)\text{ or }\left(a>2\pi n_{1}+\frac{\pi }{2}\text{ and }a\leq 2\pi n_{1}+\pi \right)\right)$$

Ekuazio irrazionalak ebazteko urratsak

Trukatu aldeak, aldagaiak ezkerraldean egon daitezen.

$$\sqrt{b}\left(1-\left(\sin(a)\right)^{2}\right)=\sin(a)$$

Erabili banaketa-propietatea \sqrt{b} eta 1-\left(\sin(a)\right)^{2} biderkatzeko.

$$\sqrt{b}-\sqrt{b}\left(\sin(a)\right)^{2}=\sin(a)$$

Konbinatu b duten gai guztiak.

$$\left(1-\left(\sin(a)\right)^{2}\right)\sqrt{b}=\sin(a)$$

Zatitu ekuazioaren bi aldeak 1-\left(\sin(a)\right)^{2} balioarekin.

$$\frac{\left(-\left(\sin(a)\right)^{2}+1\right)\sqrt{b}}{-\left(\sin(a)\right)^{2}+1}=\frac{\sin(a)}{-\left(\sin(a)\right)^{2}+1}$$

1-\left(\sin(a)\right)^{2} balioarekin zatituz gero, 1-\left(\sin(a)\right)^{2} balioarekiko biderketa desegiten da.

$$\sqrt{b}=\frac{\sin(a)}{-\left(\sin(a)\right)^{2}+1}$$

Zatitu \sin(a) balioa 1-\left(\sin(a)\right)^{2} balioarekin.

$$\sqrt{b}=\frac{\tan(a)}{\cos(a)}$$

Egin ekuazioaren bi aldeak ber bi.

$$b=\frac{\left(\tan(a)\right)^{2}}{\left(\cos(a)\right)^{2}}$$

Ebatzi: a

$$\left\{\begin{matrix}a=2\pi n_{2}+\arcsin(\frac{\sqrt{4b+1}-1}{2\sqrt{b}})\text{, }n_{2}\in \mathrm{Z}\text{; }a=2\pi n_{3}-\arcsin(\frac{\sqrt{4b+1}-1}{2\sqrt{b}})+\pi \text{, }n_{3}\in \mathrm{Z}\text{, }&b>0\\a=\pi n_{1}\text{, }n_{1}\in \mathrm{Z}\text{, }&b=0\end{matrix}\right.$$